15
$\begingroup$

Suppose that, for some integer $k$, a series $f(q) \in \mathbb Q \otimes \mathbb Z[[q]]$ has the property that for every prime $p$, $f(q)$ is the $q$-expansion of a $p$-adic modular form of weight $k$ and level $1$. Write $\mathcal M_k$ for the $\mathbb Q$-vector space of such power series, and $\mathcal M$ for the graded sum of the $\mathcal M_k$'s, which has the structure of a graded $\mathbb Q$-algebra.

Every classical modular form of weight $k$ defined over $\mathbb Q$ yields an $f \in \mathcal M_k$, but it turns out that there exist nontrivial examples of "everywhere $p$-adic" modular forms as above, which are not classical modular forms. Serre's operator $\theta = q \frac{d}{dq}$ preserves the space of $p$-adic modular forms for every $p$ (more precisely it takes $p$-adic modular forms of weight $k$ to $p$-adic modular forms of weight $k+2$), and since $\theta$ acts on $q$-expansions in a manner independent from $p$, it preserves $\mathcal M$, even though $\theta f$ is in general not a classical modular form if $f$ is.

More generally, any nearly holomorphic modular form defined over $\mathbb Q$ determines an element of $\mathcal M$, via the map which takes a nearly holomorphic modular form $f_0(\tau) + f_1(\tau)/y + \dots + f_r(\tau)/y^r$ to its "constant term" $f_0$. The function $f_0$ is called a quasimodular form in the terminology of Zagier and Kaneko. Urban shows that the $q$-expansion of $f_0$ is a $p$-adic modular form for all $p$, but $f_0$ is not a classical modular form in general. For instance, the quasimodular Eisenstein series $E_2(q)$ is a $p$-adic modular form of weight $2$ for every $p$, even though the space of classical modular forms of weight $2$ is zero.

I am curious to know whether there are any other examples. Does there exist an $f(q) \in \mathcal M_k$ which is not a quasimodular form? It seems quite likely to me that there is not. If this were true, this would constitute a sort of local-to-global principle. But I have no idea how to attack this problem; I don't even see any reason why $\mathcal M_k$ should be finite-dimensional!

$\endgroup$
  • $\begingroup$ Same level or varying level? $\endgroup$ – Will Sawin Feb 10 '15 at 2:00
  • $\begingroup$ @Will, I am happy to have everything in level $1$. $\endgroup$ – Bruno Joyal Feb 10 '15 at 2:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.